Current location - Training Enrollment Network - Mathematics courses - Senior one requires two math problems!
Senior one requires two math problems!
I thought we were going to use vectors. . . . . We are now required to study 1, 4.

Two possibilities

( 1)

Take point B as the vertical foot and make a straight line L 1 vertical BC.

From B(6, 1)C(3, 3), the analytical formula of the straight line BC is: Y=-2X/3+5.

That is, the slope of the straight line BC is K=-2/3.

Because the straight line BC is perpendicular to the straight line L 1.

So the slope of L 1 is K=3/2.

Because after B.

So the analytical formula of the straight line L 1 is: Y=3X/2-8.

Make a straight line L2 perpendicular to L 1 through point D.

Since L2 passes through point D, the slope is K=-2/3.

So the analytical formula of L2 line is: Y=-2X/3+ 19/3.

Solve the equation L 1, L2

X=86/ 13,Y=25/ 13。

That is, the intersection of L 1 and L2, that is, point A, A(86/ 13, 25/ 13).

(2)

Make a straight line L 1 vertical DC through point d.

Because c (3,3) d (2,5)

Therefore, the analytical formula of straight line DC is: Y=-2X+9.

That is, the slope of the straight line CD is: K=-2.

Because the straight line DC is perpendicular to the straight line L 1.

So the slope of L 1 is: K= 1/2.

Because L 1 intersects with point d,

Therefore, the analytical formula of the straight line L 1 is: Y=X/2+4.

Make a straight line L2 perpendicular to L 1 through point B.

So the slope of the straight line L2 is: K=-2.

Because the straight line L2 passes through the point b,

Therefore, the analytical expression of line L2 is: rate Y=-2X+ 13.

Equation solving line L 1, L2

X= 18/5,Y=29/5。

That is, the intersection of L 1 and L2, that is, points A, A( 18/5, 29/5).